ch3 51 100

Prévia do material em texto

PROBLEM 3.51
KNOWN: Pipe wall temperature and convection conditions associated with water flow through the pipe
and ice layer formation on the inner surface.
FIND: Ice layer thickness δ.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Negligible pipe wall thermal
resistance, (3) negligible ice/wall contact resistance, (4) Constant k.
PROPERTIES: Table A.3, Ice (T = 265 K): k ≈ 1.94 W/m⋅K.
ANALYSIS: Performing an energy balance for a control surface about the ice/water interface, it follows
that, for a unit length of pipe,
conv condq q′ ′=
( )( ) ( )s,i s,oi 1 ,i s,i 2 1
T T
h 2 r T T
ln r r 2 k
π
π
∞
−
− =
Dividing both sides of the equation by r2,
( )
( ) ( )( )
s,i s,o2 1
22 1 i 2 ,i s,i
T Tln r r k 1.94 W m K 15 C 0.097
r r h r T T 3 C2000 W m K 0.05m∞
−
⋅
= × = × =
−
⋅
$
$
The equation is satisfied by r2/r1 = 1.114, in which case r1 = 0.050 m/1.114 = 0.045 m, and the ice layer
thickness is
2 1r r 0.005m 5mmδ = − = = <
COMMENTS: With no flow, hi → 0, in which case r1 → 0 and complete blockage could occur. The
pipe should be insulated.
PROBLEM 3.52
KNOWN: Inner surface temperature of insulation blanket comprised of two semi-cylindrical shells of different
materials. Ambient air conditions.
FIND: (a) Equivalent thermal circuit, (b) Total heat loss and material outer surface temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial conduction, (3) Infinite contact
resistance between materials, (4) Constant properties.
ANALYSIS: (a) The thermal circuit is,
conv,A conv,B 2R R 1/ r hπ′ ′= =
( ) ( )2 1cond A
A
ln r / r
R
 kπ
′ = <
( ) ( )2 icond
B
ln r / r
R
 kπ
′ =B
The conduction resistances follow from Section 3.3.1 and Eq. 3.28. Each resistance is larger by a factor of 2 than
the result of Eq. 3.28 due to the reduced area.
(b) Evaluating the thermal resistances and the heat rate ( )A Bq =q q ,′ ′ ′+
( ) 12convR 0.1m 25 W/m K 0.1273 m K/Wπ −′ = × × ⋅ = ⋅
( ) ( ) ( ) ( )cond A cond B cond Aln 0.1m/0.05mR 0.1103 m K/W R 8 R 0.8825 m K/W2 W/m Kπ′ ′ ′= = ⋅ = = ⋅× ⋅
( ) ( )
s,1 s,1
conv convcond A cond B
T T T T
q =
R R R R
∞ ∞− −
′ +
′ ′ ′ ′+ +
( )
( )
( )
( ) ( )
500 300 K 500 300 K
q = 842 198 W/m=1040 W/m.
0.1103+0.1273 m K/W 0.8825+0.1273 m K/W
− −
′ + = +
⋅ ⋅
<
Hence, the temperatures are
( ) ( )s,1 As,2 A cond A W m KT T q R 500K 842 0.1103 407Km W
⋅
′ ′= − = − × = <
( ) ( )s,1 Bs,2 B cond B W m KT T q R 500K 198 0.8825 325K.m W
⋅
′ ′= − = − × = <
COMMENTS: The total heat loss can also be computed from ( )s,1 equivq = T T / R ,∞′ −
where ( )( ) ( ) 11 1equiv conv,A cond(B) conv,Bcond AR R R R R 0.1923 m K/W.−− −′ ′ ′ ′= + + + = ⋅   
Hence ( )q = 500 300 K/0.1923 m K/W=1040 W/m.′ − ⋅
PROBLEM 3.53
KNOWN: Surface temperature of a circular rod coated with bakelite and adjoining fluid
conditions.
FIND: (a) Critical insulation radius, (b) Heat transfer per unit length for bare rod and for
insulation at critical radius, (c) Insulation thickness needed for 25% heat rate reduction.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in r, (3)
Constant properties, (4) Negligible radiation and contact resistance.
PROPERTIES: Table A-3, Bakelite (300K): k = 1.4 W/m⋅K.
ANALYSIS: (a) From Example 3.4, the critical radius is
cr 2
k 1.4 W/m K
r 0.01m.
h 140 W/m K
⋅
= = =
⋅
<
(b) For the bare rod,
( ) ( )i iq =h D T Tπ ∞′ −
( ) ( )2
Wq =140 0.01m 200 25 C=770 W/m
m K
π′ × −
⋅
$ <
For the critical insulation thickness,
( )
( )
( )
( )
i
cr i
2
cr
200 25 CT Tq =
ln r / r ln 0.01m/0.005m1 1
2 r h 2 k 2 1.4 W/m K2 0.01m 140 W/m Kπ π ππ
∞
−
−
′ =
+ +
× ⋅× × ⋅
$
( )
175 Cq = 909 W/m
0.1137+0.0788 m K/W
′ =
⋅
$
<
(c) The insulation thickness needed to reduce the heat rate to 577 W/m is obtained from
( )
( )
( )
( )
i
i
2
200 25 CT T Wq = 577
ln r/r ln r/0.005m m1 1
2 rh 2 k 2 1.4 W/m K2 r 140 W/m Kπ π ππ
∞
−
−
′ = =
+ +
× ⋅
⋅
$
From a trial-and-error solution, find
r ≈ 0.06 m.
The desired insulation thickness is then
( ) ( )ir r 0.06 0.005 m=55 mm.δ = − ≈ − <
PROBLEM 3.54
KNOWN: Geometry of an oil storage tank. Temperature of stored oil and environmental
conditions.
FIND: Heater power required to maintain a prescribed inner surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial
direction, (3) Constant properties, (4) Negligible radiation.
PROPERTIES: Table A-3, Pyrex (300K): k = 1.4 W/m⋅K.
ANALYSIS: The rate at which heat must be supplied is equal to the loss through the
cylindrical and hemispherical sections. Hence,
cyl hemi cyl spherq=q 2q q q+ = +
or, from Eqs. 3.28 and 3.36,
( )
s,i s,i
o i
2o i o o
T T T T
q=
ln D / D 1 1 1 1 1
2 Lk D Lh 2 k D D D hπ π π π
∞ ∞− −+  
+ − +  
( )
( ) ( ) ( )
( )
( ) ( ) ( )
2
-1
2 2
-3 -3 -3 -3
400 300 K
q= ln 1.04 1
2 2m 1.4 W/m K 1.04m 2m 10 W/m K
400 300 K
 + 1 11 0.962 m
2 1.4 W/m K 1.04m 10 W/m K
100K 100Kq=
2.23 10 K/W + 15.30 10 K/W 4.32 10 K/W + 29.43 10
π π
π π
−
+
⋅
⋅
−
− +
⋅
⋅
+
× × × ×
q = 5705W + 2963W = 8668W. <
PROBLEM 3.55
KNOWN: Diameter of a spherical container used to store liquid oxygen and properties of insulating
material. Environmental conditions.
FIND: (a) Reduction in evaporative oxygen loss associated with a prescribed insulation thickness, (b)
Effect of insulation thickness on evaporation rate.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Negligible conduction resistance of
container wall and contact resistance between wall and insulation, (3) Container wall at boiling point of
liquid oxygen.
ANALYSIS: (a) Applying an energy balance to a control surface about the insulation, in outE E−� � = 0, it
follows that conv rad condq q q q+ = = . Hence,
s,2 sur s,2 s,2 s,1
t,conv t,rad t,cond
T T T T T T
q
R R R
∞
− − −
+ = = (1)
where ( ) 12t,conv 2R 4 r hπ −= , ( ) 12t,rad 2 rR 4 r hπ −= , ( ) ( ) ( )[ ]t,cond 1 2R 1 4 k 1 r 1 rπ= − , and, from Eq.
1.9, the radiation coefficient is ( )( )2 2r s,2 sur s,2 surh T T T Tεσ= + + . With t = 10 mm (r2 = 260 mm), ε =
0.2 and T∞ = Tsur = 298 K, an iterative solution of the energy balance equation yields Ts,2 ≈ 297.7 K,
where Rt,conv = 0.118 K/W, Rt,rad = 0.982 K/W and Rt,cond = 76.5 K/W. With the insulation, it follows that
the heat gain is
qw ≈ 2.72 W
Without the insulation, the heat gain is
s,1 sur s,1
wo
t,conv t,rad
T T T T
q
R R
∞
− −
= +
where, with r2 = r1, Ts,1 = 90 K, Rt,conv = 0.127 K/W and Rt,rad = 3.14 K/W. Hence,
qwo = 1702 W
With the oxygen mass evaporation rate given by �m = q/hfg, the percent reduction in evaporated oxygen is
wo w wo w
wo wo
m m q q
% Re duction 100% 100%
m q
− −
= × = ×
� �
�
Hence,
( )1702 2.7 W% Re duction 100% 99.8%
1702 W
−
= × = <
Continued...
PROBLEM 3.55 (Cont.)
(b) Using Equation (1) to compute Ts,2 and q as a function of r2, the corresponding evaporation rate, �m =
q/hfg, may be determined. Variations of q and �m with r2 are plotted as follows.
0.25 0.26 0.27 0.28 0.29 0.3
Outer radius of insulation, r2(m)
0.1
1
10
100
1000
10000
H
ea
t g
ai
n,
 q
(W
)
 
0.25 0.26 0.27 0.28 0.29 0.3
Outer radius of insulation, r2(m)
1E-6
1E-5
0.0001
0.001
0.01
Ev
ap
or
at
io
n 
ra
te
, m
do
t(k
g/s
)
Because of its extremely low thermal conductivity, significant benefits are associated with using even a
thin layer of insulation. Nearlythree-order magnitude reductions in q and �m are achieved with r2 = 0.26
m. With increasing r2, q and �m decrease from values of 1702 W and 8×10-3 kg/s at r2 = 0.25 m to 0.627
W and 2.9×10-6 kg/s at r2 = 0.30 m.
COMMENTS: Laminated metallic-foil/glass-mat insulations are extremely effective and corresponding
conduction resistances are typically much larger than those normally associated with surface convection
and radiation.
PROBLEM 3.56
KNOWN: Sphere of radius ri, covered with insulation whose outer surface is exposed to a
convection process.
FIND: Critical insulation radius, rcr.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (spherical)
conduction, (3) Constant properties, (4) Negligible radiation at surface.
ANALYSIS: The heat rate follows from the thermal circuit shown in the schematic,
( )i totq= T T / R∞−
where tot t,conv t,condR R R and= +
t,conv 2
s
1 1R
hA 4 hrπ
= = (3.9)
t,cond
i
1 1 1R
4 k r rπ
 
= −  
(3.36)
If q is a maximum or minimum, we need to find the condition for which
totd R 0.
dr
=
It follows that
2 2 3i
d 1 1 1 1 1 1 1 1 0
dr 4 k r r 4 k 2 h4 hr r rπ π ππ
    
− + = + − =       
giving
cr
k
r 2
h
=
The second derivative, evaluated at r = rcr, is
cr
tot
3 4
r=r
dRd 1 1 3 1
 
dr dr 2 k 2 hr rπ π
 
= − +    
( ) ( )3 3
1 1 3 1 1 1 31 0
2 k 2 h 2k/h 2 k 22k/h 2k/hπ π π
   
= − + = − + >      
Hence, it follows no optimum Rtot exists. We refer to this condition as the critical insulation
radius. See Example 3.4 which considers this situation for a cylindrical system.
PROBLEM 3.57
KNOWN: Thickness of hollow aluminum sphere and insulation layer. Heat rate and inner
surface temperature. Ambient air temperature and convection coefficient.
FIND: Thermal conductivity of insulation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)
Constant properties, (4) Negligible contact resistance, (5) Negligible radiation exchange at
outer surface.
PROPERTIES: Table A-1, Aluminum (523K): k ≈ 230 W/m⋅K.
ANALYSIS: From the thermal circuit,
( )
( ) ( )( )
1 1
2 31 2tot
2A1 I 3
2I
T T T Tq= 1/ r 1/ r1/r 1/ r 1R
4 k 4 k h4 r
250 20 C
q= 80 W
1/0.15 1/ 0.18 1/ 0.18 1/ 0.30 1 K
4 230 4 k W30 4 0.3
π π π
π π π
∞ ∞− −
=
−−
+ +
−
= 
− − + +  
$
or
4
I
0.177 2303.84 10 0.029 2.875.
k 80
−× + + = =
Solving for the unknown thermal conductivity, find
kI = 0.062 W/m⋅K. <
COMMENTS: The dominant contribution to the total thermal resistance is made by the
insulation. Hence uncertainties in knowledge of h or kA1 have a negligible effect on the
accuracy of the kI measurement.
PROBLEM 3.58
KNOWN: Dimensions of spherical, stainless steel liquid oxygen (LOX) storage container. Boiling
point and latent heat of fusion of LOX. Environmental temperature.
FIND: Thermal isolation system which maintains boil-off below 1 kg/day.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible thermal resistances
associated with internal and external convection, conduction in the container wall, and contact between
wall and insulation, (3) Negligible radiation at exterior surface, (4) Constant insulation thermal
conductivity.
PROPERTIES: Table A.1, 304 Stainless steel (T = 100 K): ks = 9.2 W/m⋅K; Table A.3, Reflective,
aluminum foil-glass paper insulation (T = 150 K): ki = 0.000017 W/m⋅K.
ANALYSIS: The heat gain associated with a loss of 1 kg/day is
( )5fg 1kg dayq mh 2.13 10 J kg 2.47 W86, 400s day= = × =�
With an overall temperature difference of ( )bpT T∞ − = 150 K, the corresponding total thermal
resistance is
tot
T 150 K
R 60.7 K W
q 2.47 W
∆
= = =
Since the conduction resistance of the steel wall is
( )
3
t,cond,s
s 1 2
1 1 1 1 1 1
R 2.4 10 K W
4 k r r 4 9.2 W m K 0.35 m 0.40 mπ π
−
= − = − = ×
⋅
        
it is clear that exclusive reliance must be placed on the insulation and that a special insulation of very low
thermal conductivity should be selected. The best choice is a highly reflective foil/glass matted
insulation which was developed for cryogenic applications. It follows that
( )t,cond,i i 2 3 3
1 1 1 1 1 1
R 60.7 K W
4 k r r 4 0.000017 W m K 0.40 m rπ π
= = − = −
⋅
         
which yields r3 = 0.4021 m. The minimum insulation thickness is therefore δ = (r3 - r2) = 2.1 mm.
COMMENTS: The heat loss could be reduced well below the maximum allowable by adding more
insulation. Also, in view of weight restrictions associated with launching space vehicles, consideration
should be given to fabricating the LOX container from a lighter material.
PROBLEM 3.59
KNOWN: Diameter and surface temperature of a spherical cryoprobe. Temperature of surrounding
tissue and effective convection coefficient at interface between frozen and normal tissue.
FIND: Thickness of frozen tissue layer.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Negligible contact resistance
between probe and frozen tissue, (3) Constant properties.
ANALYSIS: Performing an energy balance for a control surface about the phase front, it follows that
conv condq q 0− =
Hence,
( )( ) ( ) ( )[ ]s,2 s,122 s,2 1 2
T T
h 4 r T T
1 r 1 r 4 k
π
π
∞
−
− =
−
( ) ( )[ ] ( )( )
s,2 s,12
2 1 2
s,2
T Tk
r 1 r 1 r
h T T
∞
−
− =
−
( )
( ) ( )( )
s,2 s,12 2
21 1 1 s,2
T Tr r k 1.5 W m K 30
1
r r hr 37T T 50 W m K 0.0015m∞
−
⋅
− = =
−
⋅
                 
2 2
1 1
r r
1 16.2
r r
− =
            
( )2 1r r 4.56=
It follows that r2 = 6.84 mm and the thickness of the frozen tissue is
2 1r r 5.34 mmδ = − = <
PROBLEM 3.60
KNOWN: Inner diameter, wall thickness and thermal conductivity of spherical vessel containing
heat generating medium. Inner surface temperature without insulation. Thickness and thermal
conductivity of insulation. Ambient air temperature and convection coefficient.
FIND: (a) Thermal energy generated within vessel, (b) Inner surface temperature of vessel with
insulation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional, radial conduction, (3) Constant properties,
(4) Negligible contact resistance, (5) Negligible radiation.
ANALYSIS: (a) From an energy balance performed at an instant for a control surface about the
pharmaceuticals, gE q,=� in which case, without the insulation
( )
( ) ( )
s,1
g
222
w 1 2 2
T T 50 25 C
E q
1 1 1 11 1 1 1
4 17 W / m K 0.50m 0.51m4 k r r 4 0.51m 6 W / m K4 r h ππ ππ
∞
−
− °
= = =
− +
− +
⋅
⋅
        
�
( )g 4 2
25 CE q 489 W
1.84 10 5.10 10 K / W− −
°
= = =
× + ×
� <
(b) With the insulation,
s,1 2
w 1 2 i 2 3 3
1 1 1 1 1 1 1T T q
4 k r r 4 k r r 4 r hπ π π
∞
    = + − + − +       
( ) ( )
4
s,1 2
1 1 1 1 KT 25 C 489 W 1.84 10
4 0.04 0.51 0.53 W4 0.53 6π π
−
   = ° + × + − +    
4
s,1
KT 25 C 489 W 1.84 10 0.147 0.047 120 C
W
− 
= ° + × + + = °   <
COMMENTS: The thermal resistance associated with the vessel wall is negligible, and without the
insulation the dominant resistance is due to convection. The thermal resistance of the insulation is
approximately three times that due to convection.
PROBLEM 3.61
KNOWN: Spherical tank of 1-m diameter containing an exothermic reaction and is at 200°C when
the ambient air is at 25°C. Convection coefficient on outer surface is 20 W/m2⋅K.
FIND: Determine the thickness of urethane foam required to reduce the exterior temperature to 40°C.
Determinethe percentage reduction in the heat rate achieved using the insulation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction
through the insulation, (3) Convection coefficient is the same for bare and insulated exterior surface,
and (3) Negligible radiation exchange between the insulation outer surface and the ambient
surroundings.
PROPERTIES: Table A-3, urethane, rigid foam (300 K): k = 0.026 W/m⋅K.
ANALYSIS: (a) The heat transfer situation for the heat rate from the tank can be represented by the
thermal circuit shown above. The heat rate from the tank is
t
cd cv
T Tq
R R
∞−
=
+
where the thermal resistances associated with conduction within the insulation (Eq. 3.35) and
convection for the exterior surface, respectively, are
( ) ( ) ( )t o o o
cd
1/ r 1/ r 1/ 0.5 1/ r 1/ 0.5 1/ r
R K / W
4 k 4 0.026 W / m K 0.3267π π
− − −
= = =
× ⋅
3 2
cv o2 2 2
s o o
1 1 1R 3.979 10 r K / W
hA 4 hr 4 20 W / m K rπ π
− −
= = = = ×
× ⋅ ×
To determine the required insulation thickness so that To = 40°C, perform an energy balance on the o-
node.
t o o
cd cv
T T T T 0
R R
∞− −+ =
( )
( )
( )
3 2
o o
200 40 K 25 40 K
0
1/ 0.5 1/ r / 0.3267 K / W 3.979 10 r K / W−
− −
+ =
− ×
( )o o ir 0.5135 m t r r 0.5135 0.5000 m 13.5 mm= = − = − = <
From the rate equation, for the bare and insulated surfaces, respectively,
( )t
o 2
t
200 25 KT Tq 10.99 kW
0.01592 K / W1/ 4 hrπ
∞
−
−
= = =
( )
( )
t
ins
cd cv
200 25T Tq 0.994 kW
R R 0.161 0.01592 K / W
∞
−
−
= = =
+ +
Hence, the percentage reduction in heat loss achieved with the insulation is,
ins o
o
q q 0.994 10.99100 100 91%
q 10.99
− −
× = − × = <
PROBLEM 3.62
KNOWN: Dimensions and materials used for composite spherical shell. Heat generation
associated with stored material.
FIND: Inner surface temperature, T1, of lead (proposal is flawed if this temperature exceeds
the melting point).
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant
properties at 300K, (4) Negligible contact resistance.
PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, MP = 601K; St.St.: 15.1 W/m⋅K.
ANALYSIS: From the thermal circuit, it follows that
31
1
tot
T T 4q= q r
R 3
π∞
−  
=   �
Evaluate the thermal resistances,
( )Pb 1 1R 1/ 4 35.3 W/m K 0.00150 K/W0.25m 0.30mπ
  = × ⋅ − =    
( )St.St. 1 1R 1/ 4 15.1 W/m K 0.000567 K/W0.30m 0.31mπ
  = × ⋅ − =    
( )2 2 2convR 1/ 4 0.31 m 500 W/m K 0.00166 K/Wπ = × × ⋅ =  
totR 0.00372 K/W.=
The heat rate is ( )( )35 3q=5 10 W/m 4 / 3 0.25m 32,725 W.π× = The inner surface
temperature is
( )1 totT T R q=283K+0.00372K/W 32,725 W∞= +
1T 405 K < MP = 601K.= <
Hence, from the thermal standpoint, the proposal is adequate.
COMMENTS: In fabrication, attention should be given to maintaining a good thermal
contact. A protective outer coating should be applied to prevent long term corrosion of the
stainless steel.
PROBLEM 3.63
KNOWN: Dimensions and materials of composite (lead and stainless steel) spherical shell used to store
radioactive wastes with constant heat generation. Range of convection coefficients h available for
cooling.
FIND: (a) Variation of maximum lead temperature with h. Minimum allowable value of h to maintain
maximum lead temperature at or below 500 K. (b) Effect of outer radius of stainless steel shell on
maximum lead temperature for h = 300, 500 and 1000 W/m2⋅K.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) Constant properties
at 300 K, (4) Negligible contact resistance.
PROPERTIES: Table A-1, Lead: k = 35.3 W/m⋅K, St. St.: 15.1 W/m⋅K.
ANALYSIS: (a) From the schematic, the maximum lead temperature T1 corresponds to r = r1, and from
the thermal circuit, it may be expressed as
1 totT T R q∞= +
where q = ( ) ( )( )33 5 31q 4 3 r 5 10 W m 4 3 0.25m 32,725 Wπ π= × =� . The total thermal resistance is
tot cond,Pb cond,St.St convR R R R= + +
where expressions for the component resistances are provided in the schematic. Using the Resistance
Network model and Thermal Resistance tool pad of IHT, the following result is obtained for the variation
of T1 with h.
100 200 300 400 500 600 700 800 900 1000
Convection coefficient, h(W/m^2.K)
300
400
500
600
700
M
ax
im
um
 P
b 
Te
m
pe
ra
tu
re
, T
(r1
) (
K)
T_1
Continued...
PROBLEM 3.63 (Cont.)
To maintain T1 below 500 K, the convection coefficient must be maintained at
h ≥ 181 W/m2⋅K <
(b) The effect of varying the outer shell radius over the range 0.3 ≤ r3 ≤ 0.5 m is shown below.
0.3 0.35 0.4 0.45 0.5
Outer radius of steel shell, r3(m)
350
400
450
500
550
600
M
ax
im
um
 P
b 
te
m
pe
ra
tu
re
, T
(r1
) (K
)
h = 300 W/m^2.K
h = 500 W/m^2.K
h = 1000 W/m^2.k
For h = 300, 500 and 1000 W/m2⋅K, the maximum allowable values of the outer radius are r3 = 0.365,
0.391 and 0.408 m, respectively.
COMMENTS: For a maximum allowable value of T1 = 500 K, the maximum allowable value of the
total thermal resistance is Rtot = (T1 - T∞)/q, or Rtot = (500 - 283)K/32,725 W = 0.00663 K/W. Hence, any
increase in Rcond,St.St due to increasing r3 must be accompanied by an equivalent reduction in Rconv.
PROBLEM 3.64
KNOWN: Representation of the eye with a contact lens as a composite spherical system subjected to
convection processes at the boundaries.
FIND: (a) Thermal circuits with and without contact lens in place, (b) Heat loss from anterior
chamber for both cases, and (c) Implications of the heat loss calculations.
SCHEMATIC:
r1=10.2mm k1=0.35 W/m⋅K
r2=12.7mm k2=0.80 W/m⋅K
r3=16.5mm
T∞,i=37°C hi=12 W/m
2
⋅K
T∞,o=21°C ho=6 W/m
2
⋅K
ASSUMPTIONS: (1) Steady-state conditions, (2) Eye is represented as 1/3 sphere, (3) Convection
coefficient, ho, unchanged with or without lens present, (4) Negligible contact resistance.
ANALYSIS: (a) Using Eqs. 3.9 and 3.36 to express the resistance terms, the thermal circuits are:
Without lens: <
With lens: <
(b) The heat losses for both cases can be determined as q = (T∞,i - T∞,o)/Rt, where Rt is the
thermal resistance from the above circuits.
Without lens: ( )t,wo 2 32 -3
3 3 1 1 1
R m
4 0.35 W/m K 10.2 12.7 1012W/m K4 10.2 10 m
π
π
−
= + −
× ⋅
⋅ ×
   
( )22 -3
3
191.2 K/W+13.2 K/W+246.7 K/W=451.1 K/W
6 W/m K4 12.7 10 mπ
+ =
⋅ ×
With lens: t,w 3
3 1 1 1
R 191.2 K/W+13.2 K/W+ m
4 0.80 W/m K 12.7 16.5 10π −
= −
× ⋅
   
( )22 -3
3
191.2 K/W+13.2 K/W+5.41 K/W+146.2 K/W=356.0 K/W
6W/m K4 16.5 10 mπ
+ =
⋅ ×
Hence the heat loss rates from the anterior chamber are
Without lens: ( )woq 37 21 C/451.1 K/W=35.5mW= − $ <
With lens: ( )wq 37 21 C/356.0 K/W=44.9mW= − $ <
(c) The heat loss from the anterior chamber increases by approximately 20% when the contact
lens is in place, implying that the outer radius, r3, is less than the critical radius.
PROBLEM 3.65
KNOWN: Thermal conductivity and inner and outer radii of a hollow sphere subjected to a
uniform heat flux at its outer surface and maintained at a uniform temperature on the inner
surface.
FIND: (a) Expression for radial temperature distribution, (b) Heat flux required to maintain
prescribed surface temperatures.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)
No generation, (4) Constant properties.
ANALYSIS: (a) For the assumptions, the temperature distribution may be obtained by
integrating Fourier’s law, Eq. 3.33. That is,
( )r
r1 s,1 1
r Tr r
s,12r T
q dr q 1k dT or k T T .
4 4 rrπ π
= − − = − −∫ ∫
Hence,
( )rs,1
1
q 1 1T r T
4 k r rπ
 
= + −  
or, with 22 r 2q q / 4 r ,π′′ ≡
( )
2
2 2
s,1
1
q r 1 1T r T
k r r
′′  
= + −  
<
(b) Applying the above result at r2,
( ) ( )
( )
s,2 s,1 2
2 22
2 2 1
k T T 10 W/m K 50 20 C
q 3000 W/m .
1 1 11 1 0.1m r 0.1 0.05 mr r
−
⋅ −
′′ = = = −   
−
−     
$
<
COMMENTS: (1) The desired temperature distribution could also be obtained by solving
the appropriate form of the heat equation,
2d dTr 0
dr dr
 
=  
and applying the boundary conditions ( )
2
1 s,1 2
r
dTT r T and k q .
dr

′′= − =
(2) The negative sign on ′′q2 implies heat transfer in the negative r direction.
PROBLEM 3.66
KNOWN: Volumetric heat generation occurring within the cavity of a spherical shell of
prescribed dimensions. Convection conditions at outer surface.
FIND: Expression for steady-state temperature distribution in shell.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional radial conduction, (2) Steady-state conditions, (3)
Constant properties, (4) Uniform generation within the shell cavity, (5) Negligible radiation.
ANALYSIS: For the prescribed conditions, the appropriate form of the heat equation is
2d dTr 0
dr dr
 
=  
Integrate twice to obtain,
2 1
1 2
dT C
r C and T C .
dr r
= = − + (1,2)
The boundary conditions may be obtained from energy balances at the inner and outer
surfaces. At the inner surface (ri),( ) ( ) i i3 2g cond,i r r ii iE q 4/3 r q k 4 r dT/dr) dT/dr) qr / 3k.π π= = = − = −� � � (3)
At the outer surface (ro),
( )
o
2 2
cond,o o r conv o oq k4 r dT/dr) q h4 r T r Tπ π ∞ = − = = − 
( ) ( )
or o
dT/dr) h/k T r T .∞ = − −  (4)
From Eqs. (1) and (3), 31 iC qr / 3k.= − � From Eqs. (1), (2) and (4)
3 3
i i 22
oo
3 3
i i
2 2
oo
qr qrh
 C T
k 3r k3kr
qr qr
C T .
3r k3hr
∞
∞
    
− = − + −      
= − +
� �
� �
Hence, the temperature distribution is
3 3
i i
2
o o
qr qr1 1T= T .
3k r r 3hr
∞
 
− + +  
� �
<
COMMENTS: Note that � .E q q qg cond,i cond,o conv= = =
PROBLEM 3.67
KNOWN: Spherical tank of 3-m diameter containing LP gas at -60°C with 250 mm thickness of
insulation having thermal conductivity of 0.06 W/m⋅K. Ambient air temperature and convection
coefficient on the outer surface are 20°C and 6 W/m2⋅K, respectively.
FIND: (a) Determine the radial position in the insulation at which the temperature is 0°C and (b) If
the insulation is pervious to moisture, what conclusions can be reached about ice formation? What
effect will ice formation have on the heat gain? How can this situation be avoided?
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional, radial (spherical) conduction
through the insulation, and (3) Negligible radiation exchange between the insulation outer surface and
the ambient surroundings.
ANALYSIS: (a) The heat transfer situation can be represented by the thermal circuit shown above.
The heat gain to the tank is
( )
( )t 3ins cv
20 60 KT Tq 612.4 W
R R 0.1263 4.33 10 K / W
∞
−
 − −
−  
= = =
+ + ×
where the thermal resistances for the insulation (see Table 3.3) and the convection process on the
outer surface are, respectively,
( ) 1i o
ins
1/1.50 1/1.75 m1/ r 1/ rR 0.1263 K / W
4 k 4 0.06 W / m Kπ π
−
−
−
= = =
× ⋅
( )
3
cv 2 22s o
1 1 1R 4.33 10 K / W
hA h4 r 6 W / m K 4 1.75 mπ π
−
= = = = ×
⋅ ×
To determine the location within the insulation where Too (roo) = 0°C, use the conduction rate
equation, Eq. 3.35,
( )
( )
( ) 1oo t oo t
oo
i oo i
4 k T T 4 k T T1q r
1/ r 1/ r r q
π π
− − −
= = − 
−  
and substituting numerical values, find
( )( ) 1
oo
4 0.06 W / m K 0 60 K1
r 1.687 m
1.5 m 612.4 W
π
− × ⋅ − −
= − =   
<
(b) With roo = 1.687 m, we’d expect the region of the insulation ri ≤ r ≤ roo to be filled with ice
formations if the insulation is pervious to water vapor. The effect of the ice formation is to
substantially increase the heat gain since kice is nearly twice that of kins, and the ice region is of
thickness (1.687 – 1.50)m = 187 mm. To avoid ice formation, a vapor barrier should be installed at a
radius larger than roo.
PROBLEM 3.68
KNOWN: Radius and heat dissipation of a hemispherical source embedded in a substrate of
prescribed thermal conductivity. Source and substrate boundary conditions.
FIND: Substrate temperature distribution and surface temperature of heat source.
SCHEMATIC:
ASSUMPTIONS: (1) Top surface is adiabatic. Hence, hemispherical source in semi-infinite
medium is equivalent to spherical source in infinite medium (with q = 8 W) and heat transfer
is one-dimensional in the radial direction, (2) Steady-state conditions, (3) Constant properties,
(4) No generation.
ANALYSIS: Heat equation reduces to
( )
2 2
12
1 2
1 d dT
r 0 r dT/dr=C
dr drr
T r C / r+C .
 
=  
= −
Boundary conditions:
( ) ( )o sT T T r T∞∞ = =
Hence, C2 = T∞ and
( )s 1 o 1 o sT C / r T and C r T T .∞ ∞= − + = −
The temperature distribution has the form
( ) ( )s oT r T T T r / r∞ ∞= + − <
and the heat rate is
( ) ( )2 2s o o sq=-kAdT/dr k2 r T T r / r k2 r T Tπ π∞ ∞ = − − − = −  
It follows that
( )s -4o
q 4 WT T 50.9 C
k2 r 125 W/m K 2 10 mπ π
∞− = = =
⋅
$
sT 77.9 C.=
$ <
COMMENTS: For the semi-infinite (or infinite) medium approximation to be valid, the
substrate dimensions must be much larger than those of the transistor.
PROBLEM 3.69
KNOWN: Critical and normal tissue temperatures. Radius of spherical heat source and radius of tissue
to be maintained above the critical temperature. Tissue thermal conductivity.
FIND: General expression for radial temperature distribution in tissue. Heat rate required to maintain
prescribed thermal conditions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Constant k.
ANALYSIS: The appropriate form of the heat equation is
2
1 d dT
r 0
dr drr
 
=  
Integrating twice,
1
2
dT C
dr r
=
( ) 1 2CT r C
r
= − +
Since T → Tb as r → ∞, C2 = Tb. At r = ro, q = ( )
o
2
o r
k 4 r dT drπ− = 2 2o 1 o4 kr C rπ− = -4πkC1.
Hence, C1 = -q/4πk and the temperature distribution is
( ) bqT r T4 krπ= + <
It follows that
( ) bq 4 kr T r Tπ  = − 
Applying this result at r = rc,
( )( )( )q 4 0.5 W m K 0.005m 42 37 C 0.157 Wπ= ⋅ − =$ <
COMMENTS: At ro = 0.0005 m, T(ro) = ( )o bq 4 kr Tπ  +  = 92°C. Proximity of this temperature to
the boiling point of water suggests the need to operate at a lower power dissipation level.
PROBLEM 3.70
KNOWN: Cylindrical and spherical shells with uniform heat generation and surface temperatures.
FIND: Radial distributions of temperature, heat flux and heat rate.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction, (2) Uniform heat generation, (3)
Constant k.
ANALYSIS: (a) For the cylindrical shell, the appropriate form of the heat equation is
1 d dT q
r 0
r dr dr k
+ =
   
�
The general solution is
( ) 2 1 2qT r r C ln r C4k= − + +
�
Applying the boundary conditions, it follows that
( ) 21 s,1 1 1 1 2qT r T r C ln r C4k= = − + +
�
( ) 22 s,2 2 1 2 2qT r T r C ln r C4k= = − + +
�
which may be solved for
( )( ) ( ) ( )2 21 2 1 s,2 s,1 2 1C q/4k r r T T ln r /r= − + −   �
( ) 22 s,2 2 1 2C T q 4k r C ln r= + −�
Hence,
( ) ( )( ) ( )( ) ( ) ( )( )2 2 2 2 2s,2 2 2 1 s,2 s,1 2 1ln r/rT r T q 4k r r q 4k r r T T ln r /r= + − + − + −   � � <
With q k dT/dr′′ = − , the heat flux distribution is
( )
( )( ) ( )
( )
2 2
2 1 s,2 s,1
2 1
k q 4k r r T Tqq r r
2 r ln r /r
− + −
′′ = −
   �� <
Continued...
PROBLEM 3.70 (Cont.)
Similarly, with q = q′′A(r) = q′′ (2πrL), the heat rate distribution is
( )
( )( ) ( )
( )
2 2
2 1 s,2 s,12
2 1
2 Lk q 4k r r T T
q r Lqr
ln r /r
π
π
− + −
= −
   �
� <
(b) For the spherical shell, the heat equation and general solution are
2
2
1 d dT q
r 0
dr dr kr
+ =
   
�
( ) 2 1 2T(r) q 6k r C /r C= − − +�
Applying the boundary conditions, it follows that
( ) ( ) 21 s,1 1 1 1 2T r T q 6k r C /r C= = − − +�
( ) ( ) 22 s,2 2 1 2 2T r T q 6k r C /r C= = − − +�
Hence,
( )( ) ( ) ( ) ( )[ ]2 21 2 1 s,2 s,1 1 2C q 6k r r T T 1 r 1 r= − + − −   �
( ) 22 s,2 2 1 2C T q 6k r C /r= + +�
and
( ) ( )( ) ( )( ) ( ) ( ) ( )( ) ( )2 2 2 2 2s,2 2 2 1 s,2 s,1 1 21 r 1 rT r T q 6k r r q 6k r r T T 1 r 1 r−= + − − − + − −   � � <
With q′′ (r) = - k dT/dr, the heat flux distribution is
( )
( )( ) ( )
( ) ( )
2 2
2 1 s,2 s,1
21 2
q 6 r r k T Tq 1
q r r
3 1 r 1 r r
− + −
′′ = −
−
   �� <
and, with q = ( )2q 4 rπ′′ , the heat rate distribution is
( )
( )( ) ( )
( ) ( )
2 2
2 1 s,2 s,13
1 2
4 q 6 r r k T T4 q
q r r
3 1 r 1 r
ππ − + −
= −
−
   �� <
PROBLEM 3.71
KNOWN: Temperature distribution in a composite wall.
FIND: (a) Relative magnitudes of interfacial heat fluxes, (b) Relative magnitudes of thermal
conductivities, and (c) Heat flux as a function of distance x.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties.
ANALYSIS: (a) For the prescribed conditions (one-dimensional, steady-state, constant k),
the parabolic temperature distribution in C implies the existence of heat generation. Hence,
since dT/dx increases with decreasing x, the heat flux in C increases with decreasing x.
Hence,
3 4q q′′ ′′>
However, the linear temperature distributions in A and B indicate no generation, in which case
2 3q q′′ ′′=
(b) Since conservation of energy requires that 3,B 3,C B Cq q and dT/dx) dT/dx) ,′′ ′′= < it follows
from Fourier’s law that
B Ck k .>
Similarly, since 2,A 2,B A Bq q and dT/dx) dT/dx) ,′′ ′′= > it follows that
A Bk k .<
(c) It follows that the flux distribution appears as shown below.
COMMENTS: Note that, with dT/dx)4,C = 0, the interface at 4 is adiabatic.
PROBLEM 3.72
KNOWN: Plane wall with internal heat generation which is insulated at the inner surface and
subjected to a convection process at the outer surface.
FIND: Maximum temperature in the wall.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with uniform
volumetric heat generation, (3) Inner surface is adiabatic.
ANALYSIS: From Eq. 3.42, the temperature at the inner surface is given by Eq. 3.43 and is
the maximum temperature within the wall,
2
o sT qL / 2k+T .= �
The outer surface temperature follows from Eq. 3.46,
s
6 2
s 3
T T qL/h
WT 92 C+0.3 10 0.1m/500W/m K=92 C+60 C=152 C.
m
∞= +
= × × ⋅$ $ $ $
�
It follows that
( )26 3oT 0.3 10 W/m 0.1m / 2 25W/m K+152 C= × × × ⋅ $
oT 60 C+152 C=212 C.=
$ $ $ <
COMMENTS: The heat flux leaving the wall can be determined from knowledge of h, Ts
and T∞ using Newton’s law of cooling.
( ) ( )2 2conv sq h T T 500W/m K 152 92 C=30kW/m .∞′′ = − = ⋅ − $
This same result can be determined from an energy balance on the entire wall, which has the
form
g outE E 0− =� �
where
g out convE qAL and E q A.′′= = ⋅� ��
Hence,
6 3 2
convq qL=0.3 10 W/m 0.1m=30kW/m .′′ = × ×�
PROBLEM 3.73
KNOWN: Composite wall with outer surfaces exposed to convection process.
FIND: (a) Volumetric heat generation and thermal conductivity for material B required for special
conditions, (b) Plot of temperature distribution, (c) T1 and T2, as well as temperature distributions
corresponding to loss of coolant condition where h = 0 on surface A.
SCHEMATIC:
LA = 30 mm
LB = 30 mm
LC = 20 mm
kA = 25 W/m⋅K
kC = 50 W/m⋅K
ASSUMPTIONS: (1) Steady-state, one-dimensional heat transfer, (2) Negligible contact resistance at
interfaces, (3) Uniform generation in B; zero in A and C.
ANALYSIS: (a) From an energy balance on wall B,
in out g stE E E E− + =� � � �
1 2 Bq q 2qL 0′′ ′′− − + =�
( )B 1 2 Bq q q 2L′′ ′′= +� .
To determine the heat fluxes, 
„„
q1 and „„q2 , construct thermal circuits for A and C:
( ) ( )1 1 A Aq T T 1 h L k∞′′ = − + ( ) ( )2 2 C Cq T T L k 1 h∞′′ = − +
( )1 2
1 0.030 m
q 261 25 C
25 W m K1000 W m K
′′ = − +
⋅
⋅
    
$
 ( )2 2
0.020 m 1
q 211 25 C
50 W m K 1000 W m K
′′ = − +
⋅
⋅
    
$
( ) 21q 236 C 0.001 0.0012 m K W′′ = + ⋅$ ( ) 22q 186 C 0.0004 0.001 m K W′′ = + ⋅$
2
1q 107, 273 W m′′ =
2
2q 132,857 W m′′ =
Using the values for 1q′′ and 2q′′ in Eq. (1), find
( )2 6 3Bq 106,818 132,143 W m 2 0.030 m 4.00 10 W m= + × = ×� . <
To determine kB, use the general form of the temperature and heat flux distributions in wall B,
2B
1 2 x B 1
B B
q q
T(x) x C x C q (x) k x C
2k k
′′= − + + = − − +
   
� � (1,2)
there are 3 unknowns, C1, C2 and kB, which can be evaluated using three conditions,
Continued...
PROBLEM 3.73 (Cont.)
( ) ( )2BB 1 B 1 B 2
B
q
T L T L C L C
2k
− = = − − − +
�
where T1 = 261°C (3)
( ) ( )2BB 2 B 1 B 2
B
q
T L T L C L C
2k
+ = = − + + +
�
where T2 = 211°C (4)
( ) ( )Bx B 1 B B 1
B
q
q L q k L C
k
′′ ′′
− = − = − − − +
   
�
where 1q′′ = 107,273 W/m
2 (5)
Using IHT to solve Eqs. (3), (4) and (5) simultaneously with Bq� = 4.00 × 106 W/m3, find
Bk 15.3 W m K= ⋅ <
(b) Following the method of analysis in the IHT Example 3.6, User-Defined Functions, the temperature
distribution is shown in the plot below. The important features are (1) Distribution is quadratic in B, but
non-symmetrical; linear in A and C; (2) Because thermal conductivities of the materials are different,
discontinuities exist at each interface; (3) By comparison of gradients at x = -LB and +LB, find 2q′′ > 1q′′ .
(c) Using the same method of analysis as for Part (c), the temperature distribution is shown in the plot
below when h = 0 on the surface of A. Since the left boundary is adiabatic, material A will be isothermal
at T1. Find
T1 = 835°C T2 = 360°C <
-60 -40 -20 0 20 40
Wall position, x-coordinate (mm)
100
200
300
400
Te
m
pe
ra
tu
re
, T
 (C
)
T_xA, kA = 25 W/m.K
T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3
T_x, kC = 50 W/m.K
 
Loss of coolant on surface A
-60 -40 -20 0 20 40
Wall position, x-coordinate (mm)
200
400
600
800
Te
m
pe
ra
tu
re
, T
 (C
)
T_xA, kA = 25 W/m.K; adiabatic surface
T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3
T_x, kC = 50 W/m.K
PROBLEM 3.74
KNOWN: Composite wall exposed to convection process; inside wall experiences a uniform heat
generation.
FIND: (a) Neglecting interfacial thermal resistances, determine T1 and T2, as well as the heat fluxes
through walls A and C, and (b) Determine the same parameters, but consider the interfacial contact
resistances. Plot temperature distributions.
SCHEMATIC:
k W m K L mmA A= ⋅ =25 30
k W m K L mmB B= ⋅ =15 30
k W m K L mmC C= ⋅ =50 20
�q W mB –4 10
6 3
ASSUMPTIONS:
 (1) One-dimensional, steady-state heat flow, (2) Negligible contact resistance
between walls, part (a), (3) Uniform heat generation in B, zero in A and C, (4) Uniform properties, (5)
Negligible radiation at outer surfaces.
ANALYSIS:
 (a) The temperature distribution in wall B follows from Eq. 3.41,
( )
2 2
B B 2 1 1 2
2B BB
q L x T T x T T
T x 1
2k 2 L 2L
− −
= − + +
    
�
. (1)
 
The heat fluxes to theneighboring walls are found using Fourier’s law,
x
dT
q k
dx
′′ = − .
( ) ( )B 2 1B x B B B 1
B B
q T T
At x L : q L k L q
k 2L
−
′′ ′′= − − − + + =
   
� (2)
( ) ( )B 2 1B x B B B 2
B B
q T T
At x L : q L k L q
k 2L
−
′′ ′′= + − − + =
   
� (3)
 
The heat fluxes, 1q′′ and 2q′′ , can be evaluated by thermal circuits.
Substituting numerical values, find
( ) ( ) ( ) ( )21 1 A A 1q T T C 1 h L k 25 T C 1 1000 W m K 0.03m 25 W m K∞′′ = − + = − ⋅ + ⋅$ $
( ) ( ) ( )1 1 1q 25 T C 0.001 0.0012 K W 454.6 25 T′′ = − + = −$ (4)
( ) ( ) ( ) ( )22 2 C C 2q T T C 1 h L k T 25 C 1 1000 W m K 0.02 m 50 W m K∞′′ = − + = − ⋅ + ⋅$ $
( ) ( ) ( )2 2 2q T 25 C 0.001 0.0004 K W 714.3 T 25′′ = − + = −$ . (5)
Continued...
PROBLEM 3.74 (Cont.)
Substituting the expressions for the heat fluxes, Eqs. (4) and (5), into Eqs. (2) and (3), a system of two
equations with two unknowns is obtained.
Eq. (2): 6 3 2 1 1
T T
4 10 W m 0.03m 15 W m K q
2 0.03m
−
′′
− × × + ⋅ =
×
( ) ( )5 2 2 22 1 11.2 10 W m 2.5 10 T T W m 454.6 25 T− × − × − = −
1 2704.6 T 250T 131,365− = (6)
Eq. (3): 6 3 2 1 2
T T
4 10 W m 0.03m 15 W m K q
2 0.03m
−
′′+ × × − ⋅ =
×
( ) ( )5 2 2 22 1 21.2 10 W m 2.5 10 T T W m 714.3 T 25+ × − × − = −
1 2250T 964T 137,857− = − (7)
Solving Eqs. (6) and (7) simultaneously, find
T1 = 260.9°C T2 = 210.0°C <
From Eqs. (4) and (5), the heat fluxes at the interfaces and through walls A and C are, respectively,
( ) 21q 454.6 25 260.9 107, 240 W m′′ = − = − <
( ) 22q 714.3 210 25 132,146 W m′′ = − = + . <
Note directions of the heat fluxes.
(b) Considering interfacial contact resistances, we will use a different approach. The general solution for
the temperature and heat flux distributions in each of the materials is
( )A 1 2T x C x C= + x A 1q k C′′ = − ( )A B BL L x L− + ≤ ≤ − (1,2)
( ) 2BB 3 4
B
q
T x x C x C
2k
= − + +
� B
x 3
B
q
q x C
k
′′ = − +
�
B BL x L− ≤ ≤ (3,4)
( )C 5 6T x C x C= + x C 5q k C′′ = − ( )B B CL x L L+ ≤ ≤ + (5,6)
To determine C1 ... C6 and the distributions, we need to identify boundary conditions using surface
energy balances.
At x = -(LA + LB):
( )x A B cvq L L q 0′′ ′′− − − + = (7)
( ) ( )[ ]A 1 A A Bk C h T T L L∞− − + − − − (8)
At x = -LB: The heat flux must be continuous, but the temperature will be discontinuous across the
contact resistance.
( ) ( )x,A B x,B Bq L q L′′ ′′− = − (9)
( ) ( ) ( )[ ]x,A B 1A B 1B B tc,ABq L T L T L R′′ ′′− = − − − (10)
Continued...
PROBLEM 3.74 (Cont.)
At x = + LB: The same conditions apply as for x = -LB,
( ) ( )x,B B x,C Bq L q L′′ ′′+ = + (11)
( ) ( ) ( )[ ]x,B B 2B B 2C B tc,BCq L T L T L R′′ ′′+ = + − + (12)
At x = +(LB + LC):
( )x,C B C cvq L L q 0′′− + − = (13)
( ) ( )[ ]C 5 C B Ck C h T L L T 0∞− − − + − = (14)
 
Following the method of analysis in IHT Example 3.6, User-Defined Functions, we solve the system of
equations above for the constants C1 ... C6 for conditions with negligible and prescribed values for the
interfacial constant resistances. The results are tabulated and plotted below; 1q′′ and 2q′′ represent heat
fluxes leaving surfaces A and C, respectively.
Conditions T1A (°C) T1B (°C) T2B (°C) T2C (°C) 1q′′ (kW/m2) 2q′′ (kW/m2)
tcR ′′ = 0 260 260 210 210 106.8 132.0
tcR ′′ ≠ 0 233 470 371 227 94.6 144.2
-60 -40 -20 0 20 40
Wall position, x-coordinate (mm)
100
300
500
Te
m
pe
ra
tu
re
, T
 (C
)
T_xA, kA = 25 W/m.K
T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3
T_x, kC = 50 W/m.K
 
-60 -40 -20 0 20 40
Wall position, x-coordinate (mm)
100
300
500
Te
m
pe
ra
tu
re
, T
 (C
)
T_xA, kA = 25 W/m.K
T_x, kB = 15 W/m.K, qdotB = 4.00e6 W/m^3
T_x, kC = 50 W/m.K
COMMENTS: (1) The results for part (a) can be checked using an energy balance on wall B,
in out gE E E− = −� � �
1 2 B Bq q q 2L′′ ′′− = − ×�
where
2
1 2q q 107, 240 132,146 239,386 W m′′ ′′− = − − =
( )6 3 2B Bq L 4 10 W m 2 0.03 m 240, 000 W m− = − × × = −� .
Hence, we have confirmed proper solution of Eqs. (6) and (7).
(2) Note that the effect of the interfacial contact resistance is to increase the temperature at all locations.
The total heat flux leaving the composite wall (q1 + q2) will of course be the same for both cases.
PROBLEM 3.75
KNOWN: Composite wall of materials A and B. Wall of material A has uniform generation, while
wall B has no generation. The inner wall of material A is insulated, while the outer surface of
material B experiences convection cooling. Thermal contact resistance between the materials is
4 2
t,cR 10 m K / W
−
′′ = ⋅ . See Ex. 3.6 that considers the case without contact resistance.
FIND: Compute and plot the temperature distribution in the composite wall.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction with constant
properties, and (3) Inner surface of material A is adiabatic.
ANALYSIS: From the analysis of Ex. 3.6, we know the temperature distribution in material A is
parabolic with zero slope at the inner boundary, and that the distribution in material B is linear. At
the interface between the two materials, x = LA, the temperature distribution will show a
discontinuity.
( )
2 2
A
A 1A A2A A
q L x
T x 1 T 0 x L
2 k L
= − + ≤ ≤
    
�
( ) ( ) AB 1B 1B 2 A A B
B
x L
T x T T T L x L L
L
−
= − ≤ ≤ +−
Considering the thermal circuit above (see also Ex. 3.6) including the thermal contact resistance,
1A 1B 2
A
tot cond,B conv conv
T T T T T T
q q L
R R R R
∞ ∞ ∞
− − −
′′ = = = =
′′ ′′ ′′ ′′+
�
find TA(0) = 147.5°C, T1A = 122.5°C, T1B = 115°C, and T2 = 105°C. Using the foregoing equations
in IHT, the temperature distributions for each of the materials can be calculated and are plotted on the
graph below.
COMMENTS: (1) The effect of the thermal contact resistance between the materials is to increase
the maximum temperature of the system.
(2) Can you explain why the temperature distribution in the material B is not affected by the presence
of the thermal contact resistance at the materials’ interface?
Effect of thermal contact res is tance on temperature distribution
0 10 20 30 40 50 60 70
x (mm)
100
110
120
130
140
150
T 
(C
 
)
PROBLEM 3.76
KNOWN: Plane wall of thickness 2L, thermal conductivity k with uniform energy generation q.�
For case 1, boundary at x = -L is perfectly insulated, while boundary at x = +L is maintained at To =
50°C. For case 2, the boundary conditions are the same, but a thin dielectric strip with thermal
resistance 2tR 0.0005 m K / W′′ = ⋅ is inserted at the mid-plane.
FIND: (a) Sketch the temperature distribution for case 1 on T-x coordinates and describe key
features; identify and calculate the maximum temperature in the wall, (b) Sketch the temperature
distribution for case 2 on the same T-x coordinates and describe the key features; (c) What is the
temperature difference between the two walls at x = 0 for case 2? And (d) What is the location of the
maximum temperature of the composite wall in case 2; calculate this temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the plane and
composite walls, and (3) Constant properties.
ANALYSIS: (a) For case 1, the temperature distribution, T1(x) vs. x, is parabolic as shown in the
schematic below and the gradient is zero at the insulated boundary, x = -L. From Eq. 3.43,
( ) ( ) ( ) ( )
2 26 3
1 1
q 2L 5 10 W / m 2 0.020 m
T L T L 80 C
2k 2 50 W / m K
× ×
− − + = = = °
× ⋅
�
and since T1(+L) = To = 50°C, the maximum temperature occurs at x = -L,
( ) ( )1 1T L T L 80 C 130 C− = + + ° = °
(b) For case 2, the temperature distribution, T2(x) vs. x, is piece-wise parabolic,with zero gradient at
x = -L and a drop across the dielectric strip, ∆TAB. The temperature gradients at either side of the
dielectric strip are equal.
(c) For case 2, the temperature drop across the thin dielectric strip follows from the surface energy
balance shown above.
( ) ( )x AB t xq 0 T / R q 0 qL′′ ′′ ′′= ∆ = �
2 6 3
AB tT R qL 0.0005 m K / W 5 10 W / m 0.020 m 50 C.′′∆ = = ⋅ × × × = °�
(d) For case 2, the maximum temperature in the composite wall occurs at x = -L, with the value,
( ) ( )2 1 ABT L T L T 130 C 50 C 180 C− = − + ∆ = ° + ° = ° <
PROBLEM 3.77
KNOWN: Geometry and boundary conditions of a nuclear fuel element.
FIND: (a) Expression for the temperature distribution in the fuel, (b) Form of temperature
distribution for the entire system.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state conditions, (3)
Uniform generation, (4) Constant properties, (5) Negligible contact resistance between fuel
and cladding.
ANALYSIS: (a) The general solution to the heat equation, Eq. 3.39,
( )
2
2 f
d T q 0 L x +L
kdx
+ = − ≤ ≤
�
is 2 1 2
f
qT x C x+C .
2k
= − +
�
The insulated wall at x = - (L+b) dictates that the heat flux at x = - L is zero (for an energy
balance applied to a control volume about the wall, in outE E 0).= =� � Hence
( ) 1 1
f fx L
dT q qLL C 0 or C
dx k k
=−

= − − + = = −
� �
2
2
f f
q qLT x x+C .
2k k
= − −
� �
The value of Ts,1 may be determined from the energy conservation requirement that
g cond convE q q ,= =� or on a unit area basis.
( ) ( ) ( )s s,1 s,2 s,2kq 2L T T h T T .b ∞= − = −�
Hence,
( ) ( )
s,1 s,2 s,2
s
q 2 Lb q 2L
T T where T T
k h ∞
= + = +
� �
( ) ( )
s,1
s
q 2 Lb q 2L
T T .
k h ∞
= + +
� �
Continued …..
PROBLEM 3.77 (Cont.)
Hence from Eq. (1),
( ) ( ) ( ) ( )2s,1 2
s f
q Lq 2 Lb q 2 L 3T L T T C
k h 2 k∞
= = + + = − +
�
� �
which yields
2
s f
2b 2 3 LC T qL 
k h 2 k∞
 
= + + +  
�
Hence, the temperature distribution for ( )L x +L− ≤ ≤ is
2
f f s f
q qL 2b 2 3 LT x x+qL T
2k k k h 2 k ∞
 
= − − + + +  
� �
� <
(b) For the temperature distribution shown below,
( )
( )
maxL b x L: dT/dx=0, T=T
 L x +L: | dT/dx | with x
 +L x L+b: dT/dx is const.
− − ≤ ≤ −
− ≤ ≤ ↑ ↑
≤ ≤
PROBLEM 3.78
KNOWN: Thermal conductivity, heat generation and thickness of fuel element. Thickness and
thermal conductivity of cladding. Surface convection conditions.
FIND: (a) Temperature distribution in fuel element with one surface insulated and the other cooled
by convection. Largest and smallest temperatures and corresponding locations. (b) Same as part (a)
but with equivalent convection conditions at both surfaces, (c) Plot of temperature distributions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional heat transfer, (2) Steady-state, (3) Uniform generation, (4)
Constant properties, (5) Negligible contact resistance.
ANALYSIS: (a) From Eq. C.1,
( )
2 2
s,2 s,1 s,1 s,2
2f
T T T Tq L x xT x 1
2k 2 L 2L
 
− + = − + +  
� (1)
With an insulated surface at x = -L, Eq. C.10 yields
2
s,1 s,2
f
2q LT T
k
− =
� (2)
and with convection at x = L + b, Eq. C.13 yields
( ) ( )fs,2 s,2 s,1kU T T q L T T2L∞− = − −�
( ) 2s,1 s,2 s,2
f f
2LU 2q LT T T T
k k∞
− = − −
� (3)
Substracting Eq. (2) from Eq. (3),
( ) 2s,2
f f
2LU 4q L0 T T
k k∞
= − −
�
s,2
2qLT T
U∞
= +
� (4)
Continued …..
PROBLEM 3.78 (Cont.)
and substituting into Eq. (2)
s,1
f
L 1T T 2qL
k U∞
 
= + +  � (5)
Substituting Eqs. (4) and (5) into Eq. (1),
( ) 2
f f f
q qL 2 3 LT x x x qL T
2k k U 2 k ∞
 
= − − + + +  
� �
�
or, with U-1 = h-1 + b/ks,
( ) 2
f f s f
q qL 2b 2 3 LT x x x qL T
2 k k k h 2 k ∞
 
= − − + + + +  
� �
�
 (6) <
The maximum temperature occurs at x = - L and is
( )
s f
b 1 LT L 2qL T
k h k ∞
 
− = + + +  �
( ) 7 3
2
0.003m 1 0.015 m
T L 2 2 10 W / m 0.015 m 200 C 530 C
15 W / m K 60 W / m K10, 000 W / m K
− = × × × + + + ° = °
⋅ ⋅
⋅
     <
The lowest temperature is at x = + L and is
( )
2
f s f
3 qL 2b 2 3 LT L qL T 380 C
2 k k h 2 k ∞
 
+ = − + + + + = °  
�
� <
(b) If a convection condition is maintained at x = - L, Eq. C.12 reduces to
( ) ( )fs,1 s,2 s,1kU T T qL T T2L∞ − = − − −�
( ) 2s,1 s,2 s,1
f f
2 LU 2qLT T T T
k k∞
− = − −
� (7)
Subtracting Eq. (7) from Eq. (3),
( )s,2 s,1 s,1 s,2
f
2 LU0 T T T T or T T
k ∞ ∞
= − − + =
Hence, from Eq. (7)
Continued …..
PROBLEM 3.78 (Cont.)
s,1 s,2
s
qL 1 bT T T qL T
U h k∞ ∞
 
= = + = + +  
�
� (8)
Substituting into Eq. (1), the temperature distribution is
( )
2 2
2f s
qL x 1 bT x 1 qL T
2k h kL
∞
    = − + + +     
�
�
 (9) <
The maximum temperature is at x = 0 and is
( ) ( )
7 3 2
7 3
2
2 10 W / m 0.015 m 1 0.003 m
T 0 2 10 W / m 0.015 m 200 C
2 60 W / m K 15 W / m K10, 000 W / m K
×
= + × × + + °
× ⋅ ⋅
⋅
    
( )T 0 37.5 C 90 C 200 C 327.5 C= ° + ° + ° = ° <
The minimum temperature at x = ± L is
( )7 3s,1 s,2 2
1 0.003mT T 2 10 W / m 0.015m 200 C 290 C
15W / m K10,000 W / m K
  = = × + + ° = ° 
⋅
⋅ 
 <
(c) The temperature distributions are as shown.
The amount of heat generation is the same for both cases, but the ability to transfer heat from both
surfaces for case (b) results in lower temperatures throughout the fuel element.
COMMENTS: Note that for case (a), the temperature in the insulated cladding is constant and
equivalent to Ts,1 = 530°C.
-0 .0 1 5 -0 .0 0 9 -0 .0 0 3 0 .0 0 3 0 .0 0 9 0 .0 1 5
Fu e l e le m e n t lo ca tio n , x(m )
2 0 0
2 5 0
3 0 0
3 5 0
4 0 0
4 5 0
5 0 0
5 5 0
Te
m
pe
ra
tu
re
,
 
T(
C
)
In s u la te d s u rfa ce
S ym m e trica l co n ve ctio n co n d itio n s
PROBLEM 3.79
KNOWN: Wall of thermal conductivity k and thickness L with uniform generation q� ; strip heater
with uniform heat flux oq ;′′ prescribed inside and outside air conditions (hi, T∞,i, ho, T∞,o).
FIND: (a) Sketch temperature distribution in wall if none of the heat generated within the wall is lost
to the outside air, (b) Temperatures at the wall boundaries T(0) and T(L) for the prescribed condition,
(c) Value of ′′qo required to maintain this condition, (d) Temperature of the outer surface, T(L), if
oq=0 but q′′� corresponds to the value calculated in (c).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform
volumetric generation, (4) Constant properties.
ANALYSIS: (a) If none of the heat generated within the wall is
lost to the outside of the chamber, the gradient at x = 0 must be zero.
Since q� is uniform, the temperature distribution is parabolic, with
T(L) > T∞,i.
(b) To find temperatures at the boundaries of wall, begin with the
general solution to the appropriate form of the heat equation (Eq.3.40).
( ) 2 1 2qT x x C x+C2k= − +
� (1)
From the first boundary condition,
x=o 1
dT 0 C 0.
dx
= → = (2)
Two approaches are possible using different forms for the second boundary condition.
Approach No. 1: With boundary condition ( ) 1 T 0 T→ =
( ) 2 1qT x x T2k= − +
� (3)
To find T1, perform an overall energy balance on the wall
in out gE E E 0− + =� � �
( ) ( ),i 2 ,i qLh T L T qL=0 T L T T h∞ ∞ − − + = = + 
�
� (4)
Continued …..
PROBLEM 3.79 (Cont.)
and from Eq. (3) with x = L and T(L) = T2,
( )
2
2 2
1 1 2 ,i
q q qL qLT L L T or T T L T
2k 2k h 2k∞= − + = + = + +
� � � �
 (5,6)
Substituting numerical values into Eqs. (4) and (6), find
3 2
2T 50 C+1000 W/m 0.200 m/20 W/m K=50 C+10 C=60 C= × ⋅$ $ $ $ <
( )231T 60 C+1000 W/m 0.200 m / 2 4 W/m K=65 C.= × × ⋅$ $ <
Approach No. 2: Using the boundary condition
( )
x=L ,i
dTk h T L T
dx ∞
 
− = − 
yields the following temperature distribution which can be evaluated at x = 0,L for the required
temperatures,
( ) ( )2 2 ,iq qLT x x L T .2k h ∞= − − + +� �
(c) The value of oq′′ when T(0) = T1 = 65°C
follows from the circuit
1 ,o
o
o
T T
q
1/ h
∞−
′′ =
( )2 2oq 5 W/m K 65-25 C=200 W/m .′′ = ⋅ $ <
(d) With q=0,� the situation is represented
by the thermal circuit shown. Hence,
o a bq q q′′ ′′ ′′= +
1 ,o 1 ,i
o
o i
T T T T
q
1/ h L/k+1/h
∞ ∞
− −
′′ = +
which yields
1T 55 C.= $ <
PROBLEM 3.80
KNOWN: Wall of thermal conductivity k and thickness L with uniform generation and strip heater
with uniform heat flux oq′′ ; prescribed inside and outside air conditions ( ,iT∞ , hi, ,oT∞ , ho). Strip heater
acts to guard against heat losses from the wall to the outside.
FIND: Compute and plot oq′′ and T(0) as a function of q� for 200 ≤ �q ≤ 2000 W/m3 and ,iT∞ = 30, 50
and 70°C.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Uniform volumetric
generation, (4) Constant properties.
ANALYSIS: If no heat generated within the
wall will be lost to the outside of the chamber,
the gradient at the position x = 0 must be zero.
Since q� is uniform, the temperature distribution
must be parabolic as shown in the sketch.
 
To determine the required heater flux oq′′ as a function of the operation conditions q� and ,iT∞ , the
analysis begins by considering the temperature distribution in the wall and then surface energy balances
at the two wall surfaces. The analysis is organized for easy treatment with equation-solving software.
Temperature distribution in the wall, T(x): The general solution for the temperature distribution in the
wall is, Eq. 3.40,
2
1 2
q
T(x) x C x C
2k
= − + +
�
and the guard condition at the outer wall, x = 0, requires that the conduction heat flux be zero. Using
Fourier’s law,
( )x 1
x 0
dT
q (0) k kC 1dx
0 C 0
=
′′ = − = −

= = (1)
At the outer wall, x = 0,
2T(0) C= (2)
Surface energy balance, x = 0:
in outE E 0− =� �
( )o cv,oq q 0xq 0′′ ′′− ′′− = (3)
( ) ( )cv,o ,o xq h T(0) T , q 0 0∞′′ ′′= − = (4a,b) 
Continued...
PROBLEM 3.80 (Cont.)
Surface energy balance, x = L:
in outE E 0− =� �
x cv,iq (L) q 0′′ ′′− = (5)
x
x L
dT
q (L) k qL
dx
=
′′ = − = +
 � (6)
 
cv,i ,iq h T(L) T∞′′ = −  
( )2cv,i ,iqq h L T2k T 0 ∞′′ = − + −
   
� (7)
Solving Eqs. (1) through (7) simultaneously with appropriate numerical values and performing the
parametric analysis, the results are plotted below.
0 500 1000 1500 2000
Volumetric generation rate, qdot (W/m^3)
0
100
200
300
400
H
ea
te
r f
lu
x,
 q
''o
 (W
/m
^2
)
Tinfi = 30 C
Tinfi = 50 C
Tinfi = 70 C
0 500 1000 1500 2000
Volumetric generation rate, qdot (W/m^3)
20
40
60
80
100
120
W
al
l t
em
pe
ra
tu
re
, T
(0)
 (C
)
Tinfi = 30 C
Tinfi = 50 C
Tinfi = 70 C
From the first plot, the heater flux oq′′ is a linear function of the volumetric generation rate q� . As
expected, the higher q� and 
,iT∞ , the higher the heat flux required to maintain the guard condition
( xq′′ (0) = 0). Notice that for any q� condition, equal changes in ,iT∞ result in equal changes in the
required oq′′ . The outer wall temperature T(0) is also linearly dependent upon q� . From our knowledge
of the temperature distribution, it follows that for any q� condition, the outer wall temperature T(0) will
track changes in 
,iT∞ .
PROBLEM 3.81
KNOWN: Plane wall with prescribed nonuniform volumetric generation having one
boundary insulated and the other isothermal.
FIND: Temperature distribution, T(x), in terms of x, L, k, o oq and T .�
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-
direction, (3) Constant properties.
ANALYSIS: The appropriate form the heat diffusion equation is
d dT q 0.
dx dx k
 
+ =  
�
Noting that ( ) ( )oq q x q 1 x/L ,= = −� � � substitute for ( )q x� into the above equation, separate
variables and then integrate,
2
o o
1
q qdT x dT xd 1 dx x C .
dx k L dx k 2L
    
= − − = − − +          
� �
Separate variables and integrate again to obtain the general form of the temperature
distribution in the wall,
( )
2 2 3
o o
1 1 2
q qx x xdT x dx+C dx T x C x+C .
k 2L k 2 6L
   
= − − = − − +         
� �
Identify the boundary conditions at x = 0 and x = L to evaluate C1 and C2. At x = 0,
( ) ( )oo 1 2 2 oqT 0 T 0 0 C 0 C hence, C Tk= = − − + ⋅ + =
�
At x = L,
2
o o
1 1
x=L
q q LdT L0 L C hence, C
dx k 2L 2k
 
= = − − + =    
� �
The temperature distribution is
( )
2 3
o o
o
q q Lx xT x x+T .
k 2 6L 2k
 
= − − +   
� � <
COMMENTS: It is good practice to test the final result for satisfying BCs. The heat flux at
x = 0 can be found using Fourier’s law or from an overall energy balance
L
out g out o0
E E qdV to obtain q q L/2.′′= = =∫� � � �
PROBLEM 3.82
KNOWN: Distribution of volumetric heating and surface conditions associated with a quartz
window.
FIND: Temperature distribution in the quartz.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3)
Negligible radiation emission and convection at inner surface (x = 0) and negligible emission
from outer surface, (4) Constant properties.
ANALYSIS: The appropriate form of the heat equation for the quartz is obtained by
substituting the prescribed form of q� into Eq. 3.39.
( )2 o - x
2
1 qd T
e 0
kdx
αα β ′′−+ =
Integrating,
( ) ( )o - x - x
1 o 1 2
1 q 1dT
e C T q e C x+C
dx k k
α αβ β
α
′′
− −
′′= + + = − +
Boundary Conditions: ( )x=o ox=L
k dT/dx) q
k dT/dx) h T L T
β
∞
′′
− =
 − = − 
 Hence, at x = 0:
( )
o 1 o
1 o
1-
k q C q
k
C q / k
β β ′′ ′′− + =  
′′= −
At x = L:
( ) ( )
- L - L
o 1 o 1 2
1- 1-
k q e C h - q e C L+C T
k k
α αβ β
α ∞
   
′′ ′′− + = + −      
Substituting for C1 and solving for C2,
( ) ( )o 1-- L - Lo o2
qq qC 1 1 e e T .
h k k
βα αβ
α ∞
′′
′′ ′′ 
= − − + + +  
Hence, ( ) ( ) ( ) ( )o - L - x - Lo o1 q q qT x e e L x 1 1 e T .
k k h
α α αβ β
α ∞
′′
− ′′ ′′   
= − + − + − − +       <
COMMENTS: The temperature distribution depends strongly on the radiative coefficients, α
and β. For α → ∞ or β = 1, the heating occurs entirely at x = 0 (no volumetric heating).
PROBLEM 3.83
KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive
wastes. Surface convection conditions.
FIND: Radial temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties, (4) Negligible temperature drop across container wall.
ANALYSIS: The appropriate form of the heat equation is
2
o
2
o
q1 d dT q r
 r 1
r dr dr k k r
    = − = − −      
��
2 2 44
o o o
1 1 22 2
o o
q r q r q rdT qr
r C T C ln r+C .
dr 2k 4k4kr 16kr
= − + + = − + +
� � ��
From the boundary conditions,
( ) )
or=0 1 r=r o
dT dT
 | 0 C 0 k | h T r T
dr dr ∞
 = → = − = − 
2 2
o o o o o o o o
2
q r q rq r q rh C T
2 4 4k 16k ∞
 
+ − = − + + −   
� � � �
2
o o o o
2
q r 3q rC T .
4h 16k ∞
= + +
� �
Hence
( )
2 42
o o o o
o o
q r q r 3 1 r 1 rT r T .
4h k 16 4 r 16 r∞
     = + + − +        
� � <
COMMENTS: Applying the above result at ro yields
( ) ( )s o o oT T r T q r / 4h∞= = + �
The same result may be obtained by applying an energy balance to a control surface about the
container, where g convE q .=� The maximum temperature exists at r = 0.
PROBLEM 3.84
KNOWN: Cylindrical shell with uniform volumetric generation is insulated at inner surface
and exposed to convection on the outer surface.
FIND: (a) Temperature distribution in the shell in terms of i or , r , q, h, T and k,∞� (b)
Expression for the heat rate per unit length at the outer radius, ( )oq r .′
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial (cylindrical)
conduction in shell, (3) Uniform generation, (4) Constant properties.
ANALYSIS: (a) The general form of the temperature distribution and boundary conditions
are
( ) 2 1 2qT r r C ln r+C4k= − +
�
at r = ri:
i
2
i 1 1 iir
dT q 1 q0 r C 0 C r
dr 2k r 2k

= = − + + =
� �
at r = ro: ( )
o
o
r
dTk h T r T surface energy balance
dr ∞
  − = −  
2 2 2
o o o 2i i
o
q q 1 q qk r r h r r ln r C T
2k 2k r 4k 2k ∞
     
− + ⋅ = − + + −           
� � � �
2 22
o oi i
2 o
o o
qr qrr 1 rC 1 ln r T
2h r 2k 2 r ∞
         = − + + − +            
� �
Hence,
( ) ( ) 222 2 o iio
o o
qr qrq r rT r r r ln 1 T .
4k 2k r 2h r ∞
     = − + − + +        
� �� <
(b) From an overall energy balance on the shell,
( ) ( )2 2r o g o iq r E q r r .π′ ′= = −� � <
Alternatively, the heat rate may be found using Fourier’s law and the temperature distribution,
( ) ( ) ( )
o
2
2 2i
r o o o o i
or
qrdT q 1q r k 2 r 2 kr r 0 0 q r r
dr 2k 2k r
π π π
   ′ = − = − − + + + = −    
��
�
PROBLEM 3.85
KNOWN: The solid tube of Example 3.7 with inner and outer radii, 50 and 100 mm, and a thermal
conductivity of 5 W/m⋅K. The inner surface is cooled by a fluid at 30°C with a convection coefficient
of 1000 W/m2⋅K.
FIND: Calculate and plot the temperature distributions for volumetric generation rates of 1 × 105, 5
× 105, and 1 × 106 W/m3. Use Eq. (7) with Eq. (10) of the Example 3.7 in the IHT Workspace.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3) Constant
properties and (4) Uniform volumetric generation.
ANALYSIS: From Example 3.7, the temperature distribution in the tube is given by Eq. (7),
( ) ( )2 2 2 2s,2 1 22 2q q rT r T r r r n r r r4k 2k r = + − − ≤ ≤  � � � (1)
The temperature at the inner boundary, Ts,1, follows from the surface energy balance, Eq. (10),( ) ( )2 2 1 s,12 1q r r h2 r T Tπ π ∞− = −� (2)
For the conditions prescribed in the schematic with 5 3q 1 10 W / m= � , Eqs. (1) and (2), with r = r1
and T(r) = Ts,1, are solved simultaneously to find Ts,2 = 69.3°C. Eq. (1), with Ts,2 now a known
parameter, can be used to determine the temperature distribution, T(r). The results for different
values of the generation rate are shown in the graph.
Effect of generation rate on temperature distributions
50 60 70 80 90 100
Radial location, r (mm)
0
100
200
300
400
500
Te
m
pe
ra
tu
re
, T
(C
)
qdot = 1e5 W/m^3
qdot = 5e5 W/m^3
qdot = 1e6 W/m^3
COMMENTS: (1) The temperature distributions are parabolic with a zero gradient at the insulated
outer boundary, r = r2. The effect of increasing q� is to increase the maximum temperature in the
tube, which always occurs at the outer boundary.
(2) The equations used to generate the graphical result in the IHT Workspace are shown below.
// The temperature distribution, from Eq. 7, Example 3.7
T_r = Ts2 + qdot/(4*k) * (r2^2 – r^2) – qgot / (2*k) * r2^2*ln (r2/r)
// The temperature at the inner surface, from Eq. 7
Ts1 = Ts2 + qdot / (4*k) * (r2^2 – r1^2) – qdot / (2*k) * r2^2 * ln (r2/r1)
// The energy balance on the surface, from Eq. 10
pi * qdot * (r2^2 – r1^2) = h * 2 * pi * r1 * (Ts1 – Tinf)
PROBLEM 3.86
KNOWN: Diameter, resistivity, thermal conductivity, emissivity, voltage, and maximum temperature
of heater wire. Convection coefficient and air exit temperature. Temperature of surroundings.
FIND: Maximum operating current, heater length and power rating.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) Uniform wire temperature, (3) Constant properties, (4)
Radiation exchange with large surroundings.
ANALYSIS: Assuming a uniform wire temperature, Tmax = T(r = 0) ≡ To ≈ Ts, the maximum
volumetric heat generation may be obtained from Eq. (3.55), but with the total heat transfer
coefficient, ht = h + hr, used in lieu of the convection coefficient h. With
( )( ) ( ) ( )22 2 8 2 4 2 2 2r s sur s surh T T T T 0.20 5.67 10 W / m K 1473 323 K 1473 323 K 46.3 W / m Kεσ −= + + = × × ⋅ + + = ⋅
( ) 2 2th 250 46.3 W / m K 296.3W / m K= + ⋅ = ⋅
( ) ( ) ( )2 9 3tmax s
o
2 296.3W / m K2hq T T 1150 C 1.36 10 W / m
r 0.0005m∞
⋅
= − = ° = ×�
Hence, with
( )
( )
22 2 2
e ce e e
2 22c c
I L / AI R I Iq
LA A D / 4
ρ ρ ρ
π
= = = =
∀
�
( )1/ 21/ 2 22 9 3max
max 6
e
0.001mq D 1.36 10 W / mI 29.0 A
4 410 m
ππ
ρ −
   × = = =   Ω⋅   
� <
Also, with ∆E = I Re = I (ρeL/Ac),
( )
( )
2
c
6max e
110V 0.001m / 4E AL 2.98m
I 29.0A 10 m
π
ρ −
 
∆ ⋅   
= = =
Ω⋅
<
and the power rating is
( )elec maxP E I 110V 29A 3190 W 3.19kW= ∆ ⋅ = = = <
COMMENTS: To assess the validity of assuming a uniform wire temperature, Eq. (3.53) may be
used to compute the centerline temperature corresponding to maxq� and a surface temperature of
1200°C. It follows that ( )( )
9 32 2
o
o s
1.36 10 W / m 0.0005mq r
T T 1200 C 1203 C.
4 k 4 25 W / m K
×
= + = + ° = °
⋅
�
 With only a
3°C temperature difference between the centerline and surface of the wire, the assumption is
excellent.
PROBLEM 3.87
KNOWN: Energy generation in an aluminum-clad, thorium fuel rod under specified operating
conditions.
FIND: (a) Whether prescribed operating conditions are acceptable, (b) Effect of �q and h on acceptable
operating conditions.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in r-direction, (2) Steady-state conditions, (3)
Constant properties, (4) Negligible temperature gradients in aluminum and contact resistance between
aluminum and thorium.
PROPERTIES: Table A-1, Aluminum, pure: M.P. = 933 K; Table A-1, Thorium: M.P. = 2023 K, k ≈
60 W/m⋅K.
ANALYSIS: (a) System failure would occur if the melting point of either the thorium or the aluminum
were exceeded. From Eq. 3.53, the maximum thorium temperature, which exists at r = 0, is
2
o
s Th,max
qr
T(0) T T
4k
= + =
�
where, from the energy balance equation, Eq. 3.55, the surface temperature, which is also the aluminum
temperature, is
o
s Al
qr
T T T
2h∞
= + =
�
Hence,
8 3
Al s 2
7 10 W m 0.0125 m
T T 95 C 720 C 993K
14,000 W m K
× ×
= = + = =
⋅
$ $
( )28 3
Th,max
7 10 W m 0.0125m
T 993K 1449 K
4 60 W m K
×
= + =
× ⋅
<
Although TTh,max < M.P.Th and the thorium would not melt, Tal > M.P.Al and the cladding would melt
under the proposed operating conditions. The problem could be eliminated by decreasing q� , increasing
h or using a cladding material with a higher melting point.
(b) Using the one-dimensional, steady-state conduction model (solid cylinder) of the IHT software, the
following radial temperature distributions were obtained for parametric variations in q� and h.Continued...
PROBLEM 3.87 (Cont.)
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014
Radius, r(m)
800
900
1000
1100
1200
1300
1400
1500
1600
Te
m
pe
ra
tu
re
, T
(K
)
h = 10000 W/m^2.K, qdot = 7E8 W/m^3
h = 10000 W/m^2.K, qdot = 8E8 W/m^3
h = 10000 W/m^2.K, qdot = 9E9 W/m^3
 
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014
Radius, r(m)
400
600
800
1000
1200
Te
m
pe
ra
tu
re
, T
(K
)
qdot = 2E8, h = 2000 W/m^2.K
qdot = 2E8, h = 3000 W/m^2.K
qdot = 2E8, h = 5000 W/m^2.K
qdot = 2E8, h = 10000 W/m^2.K
For h = 10,000 W/m2⋅K, which represents a reasonable upper limit with water cooling, the temperature of
the aluminum would be well below its melting point for q� = 7 × 108 W/m3, but would be close to the
melting point for �q = 8 × 108 W/m3 and would exceed it for q� = 9 × 108 W/m3. Hence, under the best of
conditions, q� ≈ 7 × 108 W/m3 corresponds to the maximum allowable energy generation. However, if
coolant flow conditions are constrained to provide values of h < 10,000 W/m2⋅K, volumetric heating
would have to be reduced. Even for q� as low as 2 × 108 W/m3, operation could not be sustained for h =
2000 W/m2⋅K.
The effects of q� and h on the centerline and surface temperatures are shown below.
1E8 2.8E8 4.6E8 6.4E8 8.2E8 1E9
Energy generation, qdot (W/m^3)
0
400
800
1200
1600
2000
Ce
nt
er
lin
e 
te
m
pe
ra
tu
re
, T
(0)
 (K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
1E8 2.8E8 4.6E8 6.4E8 8.2E8 1E9
Energy generation, qdot (W/m^3)
0
400
800
1200
1600
2000
Su
rfa
ce
 te
m
pe
ra
tu
re
, T
s 
(K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
For h = 2000 and 5000 W/m2⋅K, the melting point of thorium would be approached for q� ≈ 4.4 × 108 and
8.5 × 108 W/m3, respectively. For h = 2000, 5000 and 10,000 W/m2⋅K, the melting point of aluminum
would be approached for q� ≈ 1.6 × 108, 4.3 × 108 and 8.7 × 108 W/m3. Hence, the envelope of
acceptable operating conditions must call for a reduction in q� with decreasing h, from a maximum of q�
≈ 7 × 108 W/m3 for h = 10,000 W/m2⋅K.
COMMENTS: Note the problem which would arise in the event of a loss of coolant, for which case h
would decrease drastically.
PROBLEM 3.88
KNOWN: Radii and thermal conductivities of reactor fuel element and cladding. Fuel heat generation
rate. Temperature and convection coefficient of coolant.
FIND: (a) Expressions for temperature distributions in fuel and cladding, (b) Maximum fuel element
temperature for prescribed conditions, (c) Effect of h on temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible contact
resistance, (4) Constant properties.
ANALYSIS: (a) From Eqs. 3.49 and 3.23, the heat equations for the fuel (f) and cladding (c) are
( ) ( )cf 1 1 2dT1 d dT q 1 d0 r r r 0 r r r
r dr dr r dr drfk
= − ≤ ≤ = ≤ ≤        
�
Hence, integrating both equations twice,
2
f 1 1
f 2
f f f f
dT qr C qr C
T ln r C
dr 2k k r 4k k
= − + = − + +
� � (1,2)
c 3 3
c 4
c c
dT C C
T ln r C
dr k r k
= = + (3,4)
The corresponding boundary conditions are:
) ( ) ( )f f 1 c 1r 0dT dr 0 T r T r= = = (5,6)
( )[ ]
1 1 2
c c
f c c c 2
r r r r r r
dT dTdTfk k k h T r T
dr dr dr ∞
= = =
− = − − = −
      (7,8)
Note that Eqs. (7) and (8) are obtained from surface energy balances at r1 and r2, respectively. Applying
Eq. (5) to Eq. (1), it follows that C1 = 0. Hence,
2
f 2
f
qr
T C
4k
= − +
� (9)
From Eq. (6), it follows that
2
3 11
2 4
f c
C ln rqr C C
4k k
− + = +
� (10)
Continued...
PROBLEM 3.88 (Cont.)
Also, from Eq. (7),
2
31 1
3
1
Cqr qr
or C
2 r 2
= − = −
� � (11)
Finally, from Eq. (8), 3 3 2 4
2 c
C C
h ln r C T
r k ∞
− = + −
   
 or, substituting for C3 and solving for C4
2 2
1 1
4 2
2 c
qr qrC ln r T
2r h 2k ∞
= + +
� � (12)
Substituting Eqs. (11) and (12) into (10), it follows that
2 2 2 2
1 1 1 1 1
2 2
f c 2 c
qr qr ln r qr qrC ln r T
4k 2k 2r h 2k ∞
= − + + +
� � � �
2 2 2
1 1 2 1
2
f c 1 2
qr qr r qrC ln T
4k 2k r 2r h ∞
= + +
� � � (13)
Substituting Eq. (13) into (9),
( ) 2 22 2 1 2 1f 1
f c 1 2
q qr r qr
T r r ln T
4k 2k r 2r h ∞
= − + + +
� � � (14)<
Substituting Eqs. (11) and (12) into (4),
2 2
1 2 1
c
c 2
qr r qr
T ln T
2k r 2r h ∞
= + +
� �
. (15)<
(b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2⋅K is
( ) ( ) ( )
2 28 3 8 3
f
2 10 W m 0.006 m 2 10 W m 0.006 m 0.009 m
T 0 ln
4 2 W m K 2 25 W m K 0.006 m
× × × ×
= +
× ⋅ × ⋅
 
( )
( )
28 3
2
2 10 W m 0.006 m
300 K
2 0.09 m 2000 W m K
×
+ +
× ⋅
( ) ( )fT 0 900 58.4 200 300 K 1458K= + + + = . <
(c) Temperature distributions for the prescribed values of h are as follows:
0 0.001 0.002 0.004 0.005 0.006
Radius in fuel element, r(m)
300
500
700
900
1100
1300
1500
Te
m
pe
ra
tu
re
, T
f(K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
 
0.006 0.007 0.008 0.009
Radius in cladding, r(m)
300
400
500
600
Te
m
pe
ra
tu
re
, T
c(K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
Continued...
PROBLEM 3.88 (Cont.)
Clearly, the ability to control the maximum fuel temperature by increasing h is limited, and even for h →
∞, Tf(0) exceeds 1000 K. The overall temperature drop, Tf(0) - T∞, is influenced principally by the low
thermal conductivity of the fuel material.
COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = 21 fqr 4k� = (2×108
W/m3)(0.006 m)3/8 W/m⋅K = 900 K, in which case, with no cladding and h → ∞, Tf(0) = 1200 K. To
reduce Tf(0) below 1000 K for the prescribed material, it is necessary to reduce q� .
PROBLEM 3.89
KNOWN: Dimensions and properties of tubular heater and external insulation. Internal and external
convection conditions. Maximum allowable tube temperature.
FIND: (a) Maximum allowable heater current for adiabatic outer surface, (3) Effect of internal
convection coefficient on heater temperature distribution, (c) Extent of heat loss at outer surface.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform
heat generation, (4) Negligible radiation at outer surface, (5) Negligible contact resistance.
ANALYSIS: (a) From Eqs. 7 and 10, respectively, of Example 3.7, we know that
( )2 2 22s,2 s,1 2 2 1
1
q r q
T T r ln r r
2k r 4k
− = − −
� � (1)
and
( )2 22 1
s,1 ,1
1 1
q r r
T T
2h r∞
−
= +
�
(2)
Hence, eliminating Ts,1, we obtain
( ) ( )2 2 2 2 22 2s,2 ,1 1 2 1 2
1 1 1
qr r 1 k
T T ln 1 r r 1 r r
2k r 2 h r∞
− = − − + −
   
�
Substituting the prescribed conditions (h1 = 100 W/m2⋅K),
( ) ( )4 3 3s,2 ,1T T 1.237 10 m K W q W m−∞− = × ⋅ �
Hence, with Tmax corresponding to Ts,2, the maximum allowable value of �q is
6 3
max 4
1400 400
q 8.084 10 W m
1.237 10−
−
= = ×
×
�
with
( )
2 22
e c e
2
c 2
2 1
I L A II Re
q
LA
r r
ρ ρ
π
= = =
∀
−
   
�
( ) ( ) 1/ 21/ 2 6 32 2 2 2 2max 2 1 6
e
q 8.084 10 W m
I r r 0.035 0.025 m 6406 A
0.7 10 m
π π
ρ −
×
= − = − =
× Ω⋅
          
� <
Continued …..
PROBLEM 3.89 (Cont.)
(b) Using the one-dimensional, steady-state conduction model of IHT (hollow cylinder; convection at
inner surface and adiabatic outer surface), the following temperature distributions were obtained.
0.025 0.027 0.029 0.031 0.033 0.035
Radius, r(m)
300
500
700
900
1100